 # Find all the intersection points of the following two circles: - Ausanioyck 2022-03-03 Answered
Find all the intersection points of the following two circles:
- The center circle (0,0) And its radius $\sqrt{13}$
- The center circle (1, -1) And its radius $\sqrt{17}$
(Note: Write the answer as follows $\left({x}_{1},{y}_{1}\right)\left({x}_{2},{y}_{2}\right)\right)$
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Let the equation of circle is
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$
Where, h and k are the x and y coordinates of the center of the circle.
And r is the radius of the circle.
Find the equation of circles by given information,
${\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(\sqrt{13}\right)}^{2}$
${x}^{2}+{y}^{2}=13$
${x}^{2}+{y}^{2}-13=0$ (1)
${\left(x-1\right)}^{2}+{\left(y-\left(-1\right)\right)}^{2}={\left(\sqrt{17}\right)}^{2}$
${\left(x-1\right)}^{2}+{\left(y+1\right)}^{2}=17$
${x}^{2}+1-2x+{y}^{2}+1+2y=17$
${x}^{2}+1-2x+{y}^{2}+1+2y-17=0$ (2)
Multiply equation (1) by -1 and add both the equation:
$-{x}^{2}-{y}^{2}+13=0$
$1-2x+1+2y-4=0$
$-2x+2y-2=0$
$-x+y-1=0$
$y=x+1$
Substitute the value of y in equation (1),
${x}^{2}+{y}^{2}-13=0$
${x}^{2}+{\left(x+1\right)}^{2}-13=0$
${x}^{2}+{x}^{2}+1+2x-13=0$
$2{x}^{2}+2x-12=0$
${x}^{2}+x-6=0$
${x}^{2}+3x-2x-6=0$
${x}^{2}+3x-2x-6=0$
$x\left(x+3\right)-2\left(x+3\right)=0$
$\left(x+3\right)\left(x-2\right)=0$
$x=-3,2$
${x}_{1}=-3,{x}_{2}=2$
${y}_{1}=-3+1$
${y}_{1}=-2$
${y}_{2}=2+1$
${y}_{2}=3$